Differentiability — When the Derivative Exists

• If $f$ is differentiable at $x=c$, then $f$ is continuous at $c$.
• The converse is false: $|x|$ is continuous at $0$ but not differentiable there.
• Contrapositive (useful): if $f$ is discontinuous at $c$, it cannot be differentiable at $c$.

$f$ is differentiable at $x=c$ iff the one-sided derivatives match:

• LHD $= \lim_{h\to 0^-}\dfrac{f(c+h)-f(c)}{h}$
• RHD $= \lim_{h\to 0^+}\dfrac{f(c+h)-f(c)}{h}$

Differentiable at $c$ $\iff$ LHD $=$ RHD (and both finite). In practice: differentiate each piece, then equate the two pieces' derivatives at the join (after first checking continuity there).

Split $|g(x)|$ at the points where $g(x)=0$:

• $|x|$ is continuous but not differentiable at $0$ (LHD $=-1$, RHD $=+1$).
• $|x-a|$ has its corner at $x=a$.
• $x|x|$ equals $-x^2$ for $x<0$ and $x^2$ for $x\ge 0$; both have derivative $0$ at the join, so it is differentiable at $0$ (the trap).

• Between consecutive integers, $\lfloor x\rfloor$ is constant $\Rightarrow$ derivative $=0$ there.
• At every integer it jumps (discontinuous) $\Rightarrow$ not differentiable at integers.
• A function built from $\lfloor x\rfloor$ inherits these jumps unless they cancel; check each integer in the domain.

For a piecewise $f$ differentiable at $x=c$: 1. Continuity: set the two pieces' values equal at $c$. 2. Differentiability: set the two pieces' derivatives equal at $c$. Two equations in the unknown constants — solve simultaneously.

At a point where the usual rules are unsafe, compute $f'(c)=\lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$ directly. If the limit exists (and is the same from both sides), that value is the derivative; if it doesn't, $f$ is not differentiable at $c$.